Indefinite integrals

2008-04-21 12:58 am

Let I(n,r)={x^n (a+bx)^r dx where n is a positive integer and r is real but no an integer.Prove that (n+r+1)bI(n,r)=x^n (a+bx)^r+1 - naI(n-1,r)

PS.{ <---這是In既符號

回答 (1)

LOVE

2008-04-21 4:22 am

✔ 最佳答案

I(n,r)={x^n (a+bx)^r dx

= [1/b(r+1)]{ (x^n)b(r+1)(a+bx)^r dx

= [1/b(r+1)]{ x^n d(a+bx)^(r+1)

= [1/b(r+1)][x^n (a+bx)^(r+1)] - [1/b(r+1)] { (a+bx)^(r+1) d (x^n)

= [1/b(r+1)][x^n (a+bx)^(r+1)] - [1/b(r+1)] { nx^(n-1)(a+bx)(a+bx)^r dx

= [1/b(r+1)][x^n (a+bx)^(r+1)] - [1/b(r+1)][na { x^(n-1)(a+bx)^r dx + nb{x^n(a+bx)^r dx]

= [1/b(r+1)][x^n (a+bx)^(r+1)] - [1/b(r+1)][ naI(n-1,r) + nbI(n,r)]

therefore

b(r+1)I(n,r) = [x^n (a+bx)^(r+1)] - [ naI(n-1,r) + nbI(n,r)]

b(r+1)I(n,r) = [x^n (a+bx)^(r+1)] - naI(n-1,r) - nbI(n,r)

b(r+1)I(n,r) + nbI(n,r) = x^n (a+bx)^(r+1) - naI(n-1,r)

(n+r+1)bI(n,r) = x^n (a+bx)^(r+1) - naI(n-1,r)

參考: me

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